📂Electrodynamics

Electric Field Created by a Dipole

Explanation¹

The potential due to an electric dipole $\mathbf{p}$ is as follows.

$$V_{\text{dip}}(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2} = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{p\cos\theta}{r^{2}}$$

Now, let’s assume that $\mathbf{p}$ is at the origin and parallel to the $z$ axis, as shown in the figure above. Since the electric field is the gradient of the potential, in spherical coordinates it is as follows.

$$\mathbf{E} = - \nabla V = -\left( \dfrac{\partial V}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\dfrac{\partial V}{\partial \theta}\hat{\boldsymbol{\theta}} + \dfrac{1}{r \sin\theta}\dfrac{\partial V}{\partial \phi}\hat{\boldsymbol{\phi}}\right)$$

Calculating each component gives the following.

$$\begin{align*} E_{r} &= -\dfrac{\partial V}{\partial r} = -\dfrac{1}{4\pi\epsilon_{0}}\dfrac{2p\cos\theta}{r^{3}} \\ E_{\theta} &= -\frac{1}{r}\dfrac{\partial V}{\partial \theta} = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{p\sin\theta}{r^{3}} \\ E_{\phi} &= -\dfrac{1}{r \sin\theta}\dfrac{\partial V}{\partial \phi} = 0 \end{align*}$$

Therefore, the electric field created by the dipole is as follows.

$$\begin{equation} \mathbf{E}_{\text{dip}}(r,\theta)=\frac{1}{4 \pi \epsilon_{0}}\frac{p}{r^3}(2\cos\theta \hat{\mathbf{r}} + \sin\theta \hat{\boldsymbol{\theta}}) \end{equation}$$

Formula

If we convert the above equation independently of the coordinate system, it becomes as follows.

$$\mathbf{E}_{\text{dip}}(\mathbf{r}) = \frac{1}{4 \pi \epsilon_{0}}\frac{1}{r^3}[3 (\mathbf{p} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p}]$$

Derivation

First, if we express the unit vectors of the spherical coordinate system in terms of the unit vectors of the cartesian coordinate system, it is as follows.

$$\begin{align*} \hat{\mathbf{r}} =&\ \cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\theta}} =&\ \cos\phi \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta\hat{\mathbf{y}} - \sin\theta\hat{\mathbf{z}} \end{align*}$$

Therefore, calculating the expression inside the brackets in $(1)$ gives the following.

$$\begin{align*} & 2\cos\theta \hat{\mathbf{r}} + \sin \theta \hat{\boldsymbol{\theta}} \\ =&\ 2 \cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + 2 \sin\phi \sin\theta \cos\theta \hat{\mathbf{y}} + 2 \cos^2 \theta \hat{\mathbf{z}} \\ & + \cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta \sin\theta \hat{\mathbf{y}} -\sin^2\theta \hat{\mathbf{z}} \\ =&\ 3\cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + 3 \sin\phi \sin\theta \cos\theta \hat{\mathbf{y}} + 3 \cos^2 \theta \hat{\mathbf{z}} -(\sin^2\theta + \cos^2\theta)\hat{\mathbf{z}} \\ =&\ 3 \cos\theta (\cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}}) - \hat{\mathbf{z}} \\ =&\ 3 (\hat{\mathbf{p}} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \hat{\mathbf{z}} \end{align*}$$

The last equality holds because of $\cos\theta = \hat{\mathbf{p}} \cdot \hat{\mathbf{r}}$. Now, we obtain the following result.

$$\begin{align*} \mathbf{E}_{\text{dip}}(r,\theta) =&\ \frac{1}{4 \pi \epsilon_{0}}\frac{p}{r^3}(2\cos\theta \hat{\mathbf{r}} + \sin\theta \hat{\boldsymbol{\theta}}) \\[1em] =&\ \frac{1}{4 \pi \epsilon_{0}}\frac{p}{r^3}[3 (\hat{\mathbf{p}} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \hat{\mathbf{z}}] \\[1em] =&\ \frac{1}{4 \pi \epsilon_{0}}\frac{1}{r^3}[3 (\mathbf{p} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - p \hat{\mathbf{z}}] \\[1em] =&\ \frac{1}{4 \pi \epsilon_{0}}\frac{1}{r^3}[3 (\mathbf{p} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p}] \\[1em] =&\ \mathbf{E}_{\text{dip}}( \mathbf{r} ) \end{align*}$$

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